6z^2+33z+36=0

Solution for 6z^2+33z+36=0 equation:

6z^2+33z+36=0

a = 6; b = 33; c = +36;
Δ = b2-4ac
Δ = 332-4·6·36
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-15}{2*6}=\frac{-48}{12}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+15}{2*6}=\frac{-18}{12}
=-1+1/2
$